Question

The symmetric part of the matrix $A =\begin{bmatrix} {1}&{2} &{4}\\ {6}&{8}& {2} \\ {2}&{-2}&{7}\\ \end{bmatrix} $ is

• A. $\begin{bmatrix} {1}&{4} &{3}\\ {2}&{8}& {0} \\ {3}&{0}&{7}\\ \end{bmatrix} $
• B. $\begin{bmatrix} {1}&{4} &{3}\\ {4}&{8}& {0} \\ {3}&{0}&{7}\\ \end{bmatrix} $
• C. $\begin{bmatrix} {0}&{-2} &{-1}\\ {-2}&{0}& {-2} \\ {-1}&{-2}&{0}\\ \end{bmatrix} $
• D. $\begin{bmatrix} {0}&{-2} &{1}\\ {2}&{0}& {2} \\ {-1}&{2}&{0}\\ \end{bmatrix} $

Answer 1

Answer: (B)

Given, matrix $A= \begin{bmatrix}1 & 2 & 4 \\ 6 & 8 & 2 \\ 2 & -2 & 7\end{bmatrix}$
$\therefore $ Symmetric part of $A=\frac{1}{2}\left[A+A'\right]$
$=\frac{1}{2}\left\{\begin{bmatrix}1 & 2 & 4 \\ 6 & 8 & 2 \\ 2 & -2 & 7\end{bmatrix}+ \begin{bmatrix}1 & 6 & 2 \\ 2 & 8 & -2 \\ 4 & 2 & 7\end{bmatrix}\right\}$
$=\frac{1}{2}\left\{\begin{bmatrix}2 & 8 & 6 \\ 8 & 16 & 0 \\ 6 & 0 & 14\end{bmatrix}\right\}= \begin{bmatrix}1 & 4 & 3 \\ 4 & 8 & 0 \\ 3 & 0 & 7\end{bmatrix}$